# How to Find Force on Rubber Tube from Piston System?

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1. Jul 6, 2015

### Silverlight

Hello, first of all, this isn't a homework/school related question. Long story short, my father works for a chemical plant and a few of their valve/piston/transfer system things have been breaking, which he assumes is because of too much force. With everyone in a dispute and doing their own calculations, I've been trying to help out, but being only a student, my knowledge of physics is limited. If anyone can tell me how to approach this, it would be much appreciated.

Below is a diagram he drew of the system. I cut out his calculations to focus on the diagram and so I can take a shot at it myself. I'll try to explain the picture afterwards.

Sorry. I've never used this forum before (In fact, I just made an account just for this.) but I think I attached the picture as a file. If it doesn't show up, please tell me.

So basically, it's a piston system with a rubber tube going through it. The piston uses those pinching bars (the top one pushes down, the bottom bar is stationary) to squeeze and close the tube, which can only handle so much force. There are 2 air ports: one below the piston that lets air in to keep it open, and one above on the side with the spring which lets air in when it needs to close. (Both 95psi compressed air) I'm assuming whenever the topside fills with air, there will be more pressure and push the bar down.

For the pinching bar, it's a metal rod, so pretend you took a solid cylinder of metal about 2 inches wide and pressed it down on top of a rubber tube that is 4 inches wide. How would I go about finding the force applied to the rubber tube? I don't need exact numbers or calculations, just a sense of direction on what to do. I don't need to do anything fancy like friction or whatever unless it would have a HUGE impact on the result. I'm not asking anyone to do this for me either (I guess it would be good practice for any future courses anyways , just any advice on how to approach this calculation would be great.

Once again, I'm no expert, so sorry if my explanations were confusing. If it helps, I THINK this is the link to the original manufacturer of the piston system, but it might be the wrong one:
http://redvalve.com/rv/index.php/series-5200-pneumatic [Broken]

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2. Jul 6, 2015

### 256bits

It might help if you would say what part of the valve is breaking.

Last edited by a moderator: May 7, 2017
3. Jul 6, 2015

### Silverlight

I think it's happened a few times, once it was the top bar thing snapping off and others it was the tube busting/cracking. I don't know all the details, but I just know they have been doing some "math" and coming up with different things and arguing. Whatever is going on with the system is their problem, I'm only interested in finding out who did their math right or wrong and nothing else.

I know this whole thing sounds weird, but I don't know what's going on down at the plant, so I'm just focused on solving this little physics problem. :P

4. Jul 8, 2015

### 256bits

OK. good.

The air pressure acts on the area of the piston to give a force. F = P A.
The spring force is proportional to its extension ( or compression ) F = kx

You can look up on internet for more discussion of how pressure is converted into a force. And same for a spring.
Wiki articles sometimes are a good first choice ( At other times they get carried away and deviate from the main discussion points though )

Notes:
1. the piston rod reduces the effective area of the piston in the upward direction.
( the spring contact area may have to be subtracted for the downward direction )

2. It is not evident in which direction the spring aids the movement of the piston. It might be fully extended in the down position. Or it might not.

3. Not sure what the spring load of 70 psig means. As a guess it should mean the pressure from the bottom valve necessary to bring the piston to the fully top position.